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Câu hỏi: Cho hàm số $y=f(x)$ liên tục trên đoạn [0;1] và thỏa mãn $2f(x)-f(1-x)=\sqrt{1-{{x}^{2}}},\forall x\in \mathbb{R}$. Tích phân $\int\limits_{0}^{1}{f(x)dx}$ có giá trị thuộc khoảng nào dưới đây?
A. $\left( \dfrac{1}{4};\dfrac{1}{2} \right)$
B. $\left( \dfrac{1}{2};1 \right)$
C. $\left( \dfrac{1}{8};\dfrac{1}{4} \right)$
D. $\left( 0;\dfrac{1}{4} \right)$
A. $\left( \dfrac{1}{4};\dfrac{1}{2} \right)$
B. $\left( \dfrac{1}{2};1 \right)$
C. $\left( \dfrac{1}{8};\dfrac{1}{4} \right)$
D. $\left( 0;\dfrac{1}{4} \right)$
Ta có $2f\left( x \right)-f\left( 1-x \right)=\sqrt{1-{{x}^{2}}}\Leftrightarrow \int\limits_{0}^{1}{\left[ 2f\left( x \right)-f\left( 1-x \right) \right]dx}=\int\limits_{0}^{1}{\sqrt{1-{{x}^{2}}}dx}$
$\Leftrightarrow 2\int\limits_{0}^{1}{f\left( x \right)dx}-\int\limits_{0}^{1}{f\left( 1-x \right)dx}=\dfrac{\pi }{4}\Leftrightarrow 2\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{0}^{1}{f\left( 1-x \right)d\left( 1-x \right)}=\dfrac{\pi }{4}$
$\Leftrightarrow 2\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\pi }{4}\Leftrightarrow 2\int\limits_{0}^{1}{f\left( x \right)dx}-\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\pi }{4}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\pi }{4}\in \left( \dfrac{1}{2};1 \right)$
$\Leftrightarrow 2\int\limits_{0}^{1}{f\left( x \right)dx}-\int\limits_{0}^{1}{f\left( 1-x \right)dx}=\dfrac{\pi }{4}\Leftrightarrow 2\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{0}^{1}{f\left( 1-x \right)d\left( 1-x \right)}=\dfrac{\pi }{4}$
$\Leftrightarrow 2\int\limits_{0}^{1}{f\left( x \right)dx}+\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\pi }{4}\Leftrightarrow 2\int\limits_{0}^{1}{f\left( x \right)dx}-\int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\pi }{4}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\pi }{4}\in \left( \dfrac{1}{2};1 \right)$
Đáp án B.